1. ## Help with parabolas

I have a few questions about finding equations and such for parabolas
I attempted both of these problems and would like some help finishing, or re-doing them if needed.
Find the vertex of the parabola y=2x^2-12x+23.
My Work:
y= 2x(x-6)+23
(y-23)=2x(x-6)
Am I on the right track here? I know its not done, but I don't know how to go from here assuming I did it right to begin with.

2nd problem:
Write the equation of the parabola with focus (3,4) and directrix x=1.

For the vertex I got (2,4)
So the equation so far is (x-2)^2=4P(y-4)
Help from here getting P would be great also assuming I've done the work correctly...

Thanks a bunch!

2. Originally Posted by Murphie
I have a few questions about finding equations and such for parabolas
I attempted both of these problems and would like some help finishing, or re-doing them if needed.
Find the vertex of the parabola y=2x^2-12x+23.
My Work:
y= 2x(x-6)+23
(y-23)=2x(x-6)
Am I on the right track here? I know its not done, but I don't know how to go from here assuming I did it right to begin with.
No, you are not. you need to complete the square, not factor. If you can write it as y= $2(x-a)^2+ b$, Then when x= a, y= b and, if $x\ne a$ $(x-a)^2> 0$ so y> b. The vertex is at (a,b).

2nd problem:
Write the equation of the parabola with focus (3,4) and directrix x=1.

For the vertex I got (2,4)
So the equation so far is (x-2)^2=4P(y-4)
Your vertex is correct but the equation is wrong. You have the equation for a parabola with horizontal directrix but x= 1 is vertical. Your equation should be $(y- 4)^2= 4P(x- 2)$.

Help from here getting P would be great also assuming I've done the work correctly...

Thanks a bunch!
When y= 5, say, $(5-4)^2= 1= 4P(x- 2)$, so x- 2= 1/4P and x= 2+ 1/(4P). For any parabola the distance from a point on the parabola to the focus and to the directrix must be the same. The distance from (2+ 1/(4P), 5) to (3,4) is $\sqrt{(2+ 1/(4P)- 3)^2+ (5-4)^2}= \sqrt{(-1+ 1/(4P))^2+ 1}$ and the distance to x= 1 is 2+ 1/(4P)- 1= 1+ 1/(4P). Set those equal and solve for P.

3. ## maybe this will help

Vertex

A parabola which opens up has a lowest point and a parabola which opens down has a highest point. The highest or lowest point on a parabola is called the vertex. The parabola is symmetric about a vertical line through its vertex, called the axis of symmetry. The figure below shows a parabola opening up with vertex (0.75, 0.875) and axis of symmetry x = 0.75.
Vertex Form of a Quadratic Function

To find the vertex of a parabola, we will write the function in the form
. As an example, consider the function . We first complete the square on the right side:
f(x) = 2(x2 - 4x) + 7 (factor out 2 from the terms 2x2 - 8x)
= 2(x2 - 4x + 4) + 7 - 8 (complete the square of x2 - 4x)
= 2(x - 2)2 - 1 (factor the perfect square and simplify.)
Notice that for all values of x. Thus f(x) = 2(x - 2)2 - 1 ³ -1 for all values of x and the minimum value of the function is -1 when x = 2. The point (2, -1) is the lowest point on the graph so it is the vertex of the parabola. The vertical line x = 2 is the axis of symmetry. See the graph below.

In general, is called vertex form of a quadratic function. When a quadratic function is written in vertex form, we can easily determine the vertex (h, k). If the coefficient a > 0, then the parabola opens upward and the vertex is the lowest point on the parabola. We say that k is the minimum value of the quadratic function. On the other hand, if the coefficient a < 0, then the parabola opens downward and the vertex is the highest point on the parabola. In this case, k is the maximum value of the quadratic function. Explore the role of each coefficient in the following interactive example.
Example

Write the quadratic function in vertex form. Determine the vertex and the maximum or minimum value of the function.
Solution

We will complete the square to write the function in vertex form:

The vertex form is , so the vertex is (3, -11). Since a < 0, the parabola opens downward and the vertex is the highest point. The function has a maximum value of 11. Its graph is shown below.

Once we know the vertex of a parabola, we can determine the range of the quadratic function. Consider the function, . Previously we determined that the parabola has a minimum value of -1, occurring when x = 2. Thus the range of the quadratic function is {y½ y ³ -1}. As another example, lets return to the function in the above example. The graph of this function is a parabola opening downward and the maximum value of the function is 11. Therefore, the range of the quadratic function is
Finding the Vertex Algebraically

The vertex of a quadratic function can also be determined algebraically. We first assume that the quadratic function has two x-intercepts. Then the graph is a parabola that crosses the x-axis in two distinct points. Since the parabola is symmetric with respect to a vertical line through its vertex (the axis of symmetry) the x-coordinate of the vertex is always halfway between the two x-intercepts. By the quadratic formula, the two x-intercepts are

Notice that the same number, , is being added to and subtracted from . It follows that the number is halfway between This means that the x-coordinate of the vertex is . We can then find the y-coordinate of the vertex by evaluating Although we assumed that the quadratic function had two x-interecpts when we derived our vertex formula, it also holds in the other two cases, where the parabola has one or no x-intercepts.
Example

Find the vertex of the quadratic function . Use the vertex to determine the maximum or minimum value of the function and find its range.
Solution

The vertex formula gives To find the second coordinate of the vertex, we evaluate The vertex of the parabola is (3, 53). Since a < 0, the parabola opens downward and the vertex is the highest point. This gives a maximum value of 53 and the range of the function is A graph of the function is shown below.

Recall the function, which describes the height in feet of a ball t seconds after it is thrown upward from the top of a 200 foot high building. We can now determine when the ball hits the ground and the maximum height that it reaches, as well as the time that it reaches that maximum height. When the ball hits the ground, its height above ground will be zero. This gives the quadratic equation . Using the quadratic formula, we find that the solutions are
and
(rounded to two decimal places). Since the time cannot be negative, we see that the ball strikes the ground after 5.21 seconds. The maximum height of the ball will be given by the second coordinate of the vertex and the time will be the first coordinate. Using the vertex formula we find that (rounded to tow decimal places). Next we evaluate This means that the ball reaches its maximum height of 231.64 feet after 1.41 seconds.

4. Here's the solution:

factor out the leading coefficient from the x-terms

2(x^2-6x)+23

complete the square but then subtract as well to leave the expression unchanged

2(x^2+6x+9-9)+23

Distribute the 2 to the lasterm and kick outside of the parentheses

2(x^2+6x+9)-18+23

factor the trinomial square and combine constants

2(x+3)^2+5

Now it becomes clear that this is the graph of a parabola that has shifted 3 units to the left, 5 units up, and has been "stretched" by a factor of 2.

Hence, the vertex is (-3,5)

This is not the actual solutin, but it will aid you in your effort.

5. Thanks for the help but a quick question
why did the (x^2-6x) +23 change to (x^2 +6x+9-9) +23?? I realize you completed the square but I cant understand why the -6x changed from negative to positive.
thanks so much.

6. Im sorry about the signs. Your totally right about them. But, the essence of the thing is still there.

7. No worries, thanks for the help i really appreciate it